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C ++类型转换重载

[英]C++ typecast overloading

Supposed I have a class like this: 假设我有一个这样的课:

template<class T>
class Vector2 {
public:
    Vector2(const T a, const T b) : x(a), x(b) {}
    T x;
    T y;
}

I want to be able to do something like this: 我希望能够做这样的事情:

const Vector2<double> d(31.4, 48.2);  // note the const!!!

Vector2<int> i = static_cast<Vector2<int>>(d);

// i.x == 31
// i.y == 48

I have tried overloading a generic operator but it seems to break when trying to convert from a const value. 我曾尝试重载通用运算符,但在尝试从const值转换时似乎会中断。 Help? 救命?

Provide an additional constructor that's taking another template parameter U : 提供另一个采用另一个模板参数U构造函数:

template<class T>
class Vector2 {
public:
    template <class U>
    Vector2(const Vector2<U> & other) : x(other.x), y(other.y){}

    // other code ommited
};

After all, you're trying to use Vector2<T>::Vector2(const Vector2<U> &) , where U = double and T = int . 毕竟,您尝试使用Vector2<T>::Vector2(const Vector2<U> &) ,其中U = doubleT = int

Note that this has nothing to do with your original vector being const . 请注意,这与原始向量const Instead, you're trying to construct a value of type Vector2<int> with a value of another type Value2<double> . 相反,您尝试使用另一个类型Value2<double>的值构造Vector2<int>类型的值。 Those are distinct types, and therefore you need to provide a constructor. 这些是不同的类型,因此您需要提供一个构造函数。

A possibility would be to write a cast operator which does what you want: 一种可能是编写执行所需操作的强制转换运算符:

template<class T>
class Vector2 {
public:
    Vector2(const T a, const T b) : x(a), y(b) {}
    T x;
    T y;

    template<typename U>
    operator Vector2<U>() const { return Vector2<U>( (U)x, (U)y ); }
 // ^^^^^^^^ cast operator
};

int main()
{
    const Vector2<double> d(31.4, 48.2);  // note the const!!!

    Vector2<int> i = static_cast<Vector2<int>>(d);

    return 0;
}

An additional constructor as shown in the answer of Zeta is the much more elegant solution. Zeta答案中显示的另一个构造函数是更优雅的解决方案。

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